University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 53

Answer

$\frac{2xe^{x^2}}{\sqrt {e^{x^2}}}$

Work Step by Step

$y=\int_{0}^{e^{x^2}} ({\frac{1}{\sqrt t}dt}) =\int_{0}^{e^{x^2}} (t^{-1/2}dt) =[2\sqrt t]_{0}^{e^{x^2}}=[2\sqrt {e^{x^2}}]$ Let's differentiate this! $\frac{dy}{dx}=2\cdot\frac{1}{2}({e^{x^2}})^{-1/2}\cdot(2xe^{x^2})=\frac{2xe^{x^2}}{\sqrt {e^{x^2}}}$
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